Given is a bag containing 6 red, 4 white and 8 blue balls.
By using the formula of probability we get,
P (E) = favourable outcomes / total possible outcomes
Two balls are drawn at random, so the total possible outcomes will be ${}^{18}C_3$
n (S) = 816
Let E be the event that one of the balls must be red
E = {(R) (B) (B)} or {(R) (W) (W)} or {(R) (B) (W)}
$n (E) = {}^6C_1{}^4C_1{}^8C_1+{}^6C_1{}^4C_2+{}^6C_1{}^8C_2 = 396$
P (E) = n (E) / n (S)
= 396 / 816
= 33/68